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Math Help - Working with sec...any thoughts

  1. #1
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    Working with sec...any thoughts

    Hi,

    I'm solving the integral of sec x.

    I'm using a t sub ie t = tan(x/2)

    I've integrated to ln(|1+t|/|1-t|)

    Now I believe I have to multiply inside the brackets by cos(x/2)+sin(x/2) so I'll end up with

    (cos^2(x/2)+sin^2(x/2)+2cos(x/2)sin(x/2))/cos(x)

    Could someone explain that to me?

    Thank You.
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  2. #2
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    Quote Originally Posted by minusb View Post
    Hi,

    I'm solving the integral of sec x.

    I'm using a t sub ie t = tan(x/2)

    I've integrated to ln(|1+t|/|1-t|)

    Now I believe I have to multiply inside the brackets by cos(x/2)+sin(x/2) so I'll end up with

    (cos^2(x/2)+sin^2(x/2)+2cos(x/2)sin(x/2))/cos(x)

    Could someone explain that to me?

    Thank You.
    try the substitution

    u=\sec(x)+\tan(x) \implies du = \sec(x)\tan(x)+\sec^2(x)\, dx = \sec(x) u \,dx

    This gives the integral

    \displaystyle \frac{1}{u}\,du=...
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  3. #3
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    Thanks, I know that version... The question is asking to use t=tan(x/2) substitution.
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  4. #4
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    Quote Originally Posted by minusb View Post
    I've integrated to ln(|1+t|/|1-t|)
    Well, you have done it all! Just resubstitute t and note that 1 = \tan\frac{\pi}{4}, then use the tangent addition identity:

    \displaystyle \ln\left\{\frac{1+t}{1-t}\right\}+k = \ln\left\{\frac{1+\tan\frac{x}{2}}{1-\tan\frac{x}{2}}\right\}+k = \ln\left\{\frac{\tan{\frac{\pi}{4}}+\tan\frac{x}{2  }}{\tan{\frac{\pi}{4}}-\tan\frac{x}{2}}\right\}+k = \displaystyle \ln\left\{\tan\left(\frac{\pi}{4}+\frac{x}{2}\righ  t)\right\}+k.
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  5. #5
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    Quote Originally Posted by TheCoffeeMachine View Post

     \displaystyle \ln\left\{\tan\left(\frac{\pi}{4}+\frac{x}{2}\righ  t)\right\}+k.
    Is this a common identity that is equal to secx + tanx?
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  6. #6
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    Double post...
    Last edited by minusb; November 29th 2010 at 05:25 AM.
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  7. #7
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    Quote Originally Posted by minusb View Post
    Is this a common identity that is equal to secx + tanx?
    Why do you have to show that they are equal? This is the classical integral of \displaystyle \sec{x}!
    If they have asked you to use that substitution, it should really be what they wanted!
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